我需要确保一个值在两列中是唯一的（这不是“两列”索引问题）。
Table A
Column A1 Column A2
Memphis New York -> ok
San Francisco Miami -> ok
Washington Chicago -> ok
Miami Las Vegas -> Forbidden ! Miami already exists
是否可以 ？
我的例子是城市，但不要专注于此。我真正需要的是关于生成的十六进制 ID。
在 SQL Server 中，可以借助索引视图来强制实现唯一性。您还需要在与您的Table A.
这是我的测试脚本和一些评论：
CREATE TABLE MyNumbersTable (Value int);
-- You need at least 2 rows, by the number of columns
-- you are going to implement uniqueness on
INSERT INTO MyNumbersTable
SELECT 1 UNION ALL
SELECT 2;
GO
CREATE TABLE MyUniqueCities ( -- the main table
ID int IDENTITY,
City1 varchar(50) NOT NULL,
City2 varchar(50) NOT NULL
);
GO
CREATE VIEW MyIndexedView
WITH SCHEMABINDING -- this is required for creating an indexed view
AS
SELECT
City = CASE t.Value -- after supplying the numbers table
WHEN 1 THEN u.City1 -- with the necessary number of rows
WHEN 2 THEN u.City2 -- you can extend this CASE expression
END -- to add more columns to the constraint
FROM dbo.MyUniqueCities u
INNER JOIN dbo.MyNumbersTable t
ON t.Value BETWEEN 1 AND 2 -- change here too for more columns
GO
-- the first index on an indexed view *must* be unique,
-- which suits us perfectly
CREATE UNIQUE CLUSTERED INDEX UIX_MyIndexedView ON MyIndexedView (City)
GO
-- the first two rows insert fine
INSERT INTO MyUniqueCities VALUES ('London', 'New York');
INSERT INTO MyUniqueCities VALUES ('Amsterdam', 'Prague');
-- the following insert produces an error, because of 'London'
INSERT INTO MyUniqueCities VALUES ('Melbourne', 'London');
GO
DROP VIEW MyIndexedView
DROP TABLE MyUniqueCities
DROP TABLE MyNumbersTable
GO
有用的阅读：
您需要添加一个在插入/更新后查找它的约束触发器。